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General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.

Karen C. Timberlake

Lecture Presentation

Chapter 9

Solutions


Karen C. Timberlake


Chapter 9 Solutions

A dialysis nurse informs Michelle that

• her side effects are due to her body’s inability to regulate the amount of water in her cells.

• the amount of water in her body fluids is regulated by the concentration of electrolytes and the rate at which waste products are removed from her body.


Karen C. Timberlake


Chapter 9 Readiness

Key Math Skills

• Calculating a Percentage (1.4C)

• Solving Equations (1.4D)

• Interpreting a Line Graph (1.4E)

Core Chemistry Skills

• Writing Conversion Factors from Conversion Equalities (2.5)

• Using Conversion Factors (2.6)

• Identifying Attractive Forces (6.8)

• Using Mole–Mole Factors (7.6)


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9.1 Solutions

Solutions

• are homogeneous mixtures of two or more substances.

• form when there is sufficient attraction between the solute and solvent molecules.

• have two components: the solvent, present in a larger amount, and the solute, present in a smaller amount.

Learning Goal Identify the solute and solvent in a solution;

describe the formation of a solution.


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Solutes

Solutes

• may be a liquid, gas, or solid.

• are spread evenly throughout the solution.

• mix with solvents so the solute and solvent have the same physical state.

• cannot be separated by filtration, but they can be separated by evaporation.

• are not visible, but they can give a color to the solution.


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Solutes

A solution of copper(II) sulfate (CuSO4) forms as particles of

solute dissolve and become evenly dispersed among the solvent

(water) molecules.


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Types of Solutes and Solvents

Solutes and solvents may be solids, liquids, or gases.


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Water as a Solvent

Water

• is one of the most common solvents in nature.

• is a polar molecule due to polar O–H bonds.

• molecules form hydrogen bonds important in many biological compounds.


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Formation of Solutions

Solutions form when the

solute–solvent

interactions are large

enough to overcome the

solute–solute

interactions and the

solvent–solvent

interactions.


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Solutions, Like Dissolves Like

Solutions will form when the solute and solvent have similar polarities: “like dissolves like.”


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Solutions with Ionic Solutes

NaCl crystals undergo

hydration as water molecules

surround each ion and pull it

into solution.

NaCl(s) → Na+(aq) + Cl–(aq)

solid separate ions

H2O(l)


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Solutions with Polar Solutes

A polar molecular compound such as methanol,

CH3—OH, is soluble in water because methanol has

a polar –OH group to form hydrogen bonds with water.

Polar solutes require polar solvents for a solution

to form.


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Solutions with Nonpolar Solutes

Compounds containing nonpolar molecules, such as

iodine (I2), oil, or grease, do not dissolve in water

because there are essentially no attractions between

the particles of a nonpolar solute and the polar

solvent.

Nonpolar solutes require nonpolar solvents for a

solution to form.


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Study Check

Identify the solute in each of the following solutions.

A. 2 g of sugar and 100 mL of water

B. 60.0 mL of ethyl alcohol and 30.0 mL of methyl alcohol

C. 55.0 mL of water and 1.50 g of NaCl

D. Air: 200 mL of O2 and 800 mL of N2


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Solution

Identify the solute in each of the following solutions.

A. 2 g of sugar and 100 mL of water

The solute is sugar.

B. 60.0 mL of ethyl alcohol and 30.0 mL of methyl alcohol

The solute is methyl alcohol.

C. 55.0 mL of water and 1.50 g of NaCl

The solute is NaCl.

D. Air: 200 mL of O2 and 800 mL of N2

The solute is O2.


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Study Check

When solid LiCl is added to water it dissolves because

A. the Li+ ions are attracted to the

1) oxygen atom (δ −) of water.

2) hydrogen atom (δ +) of water.

B. the Cl− ions are attracted to the

1) oxygen atom (δ −) of water.

2) hydrogen atom (δ +) of water.


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Solution

When solid LiCl is added to water it dissolves because

A. the Li+ ions are attracted to the

1) oxygen atom (σ −) of water.

B. the Cl− ions are attracted to the

2) hydrogen atom (σ +) of water.


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Study Check

Which of the following solutes will dissolve in water?

Why?

A. Na2SO4

B. gasoline (nonpolar)

C. I2

D. HCl


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Solution

Which of the following solutes will dissolve in water? Why?

Water is a polar solvent that can dissolve ionic and polar substances but not nonpolar solutes.

A. Na2SO4 will dissolve, ionic

B. gasoline (nonpolar) will not dissolve, nonpolar

C. I2 will not dissolve, nonpolar

D. HCl will dissolve, polar


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9.2 Electrolytes and Nonelectrolytes

Electrolytes in the body play an important role in maintaining the proper function of the cells and organs.

Electrolytes such as sodium, potassium, chloride, bicarbonate can be measured in a blood test.

Learning Goal Identify solutes as electrolytes or nonelectrolytes.


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Strong Electrolytes

Strong electrolytes

• dissociate 100% in water, producing positive and negative ions.

• form solutions that conduct an electric current strong enough to light a bulb.

dissociation

NaCl(s) → Na+(aq) + Cl–(aq)H2O(l)


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HF(aq) H+(aq) + F–(aq)

Weak Electrolytes

A weak electrolyte

• dissociates only slightly in water.

• forms a solution with a few

ions and mostly undissociated

molecules.

2

dissociation

recombination

H2O(l)


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Nonelectrolytes

Nonelectrolytes

• dissolve as molecules in water.

• do not produce ions in water.

• do not conduct an electric current.

C12H22O11(s) C12H22O11(aq)sucrose solution of sucrose

H2O(l)


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Solutes in Aqueous Solutions


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Study Check

Complete each for strong electrolytes in water.

1. CaCl2(s) __________

A. CaCl2(s)

B. Ca2+(aq) + Cl2−(aq)

C. Ca2+(aq) + 2Cl−(aq)

2. K3PO4(s) ___________

A. 3K+(aq) + PO43−(aq)

B. K3PO4(s)

C. K3+(aq) + P3−(aq) + O4

−(aq)

H2O(l)

H2O(l)


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Solution

Complete each for strong electrolytes in water.

1. CaCl2(s)

C. Ca2+(aq) + 2Cl−(aq)

2. K3PO4(s)

A. 3K+(aq) + PO43−(aq)

H2O(l)

H2O(l)


Karen C. Timberlake


Study Check

Which of the following reactions represents the

dissociation of a strong electrolyte in water?

A. NH3(g) + H2O(l) NH4OH(aq)

B. CH3OH(l) CH3OH(aq)

C. Na2SO4(s) 2Na+(aq) + SO42−(aq)

D. C2H5OH(l) C2H5OH (aq)

H2O(l)

H2O(l)

H2O(l)


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Solution

Which of the following reactions represents the dissociation of a strong electrolyte in water?

C. Na2SO4(s) 2Na+(aq) + SO42−(aq)

H2O(l)


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Study Check

Write the equation for the formation of a solution for

each of the following:

A. the dissociation of K2CrO4(s), a strong electrolyte,

in water

B. the partial dissociation of the weak electrolyte

H3PO4(aq) in water

C. the dissolving of the solid sugar (C12H22O11)(s)

in water


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Solution

Write the equation for the formation of a solution for each of

the following:

A. the dissociation of K2CrO4(s), a strong electrolyte, in water

K2CrO4(s) 2K+(aq) + CrO42−(aq)

B. the partial dissociation of the weak electrolyte H3PO4(aq) in water

H3PO4(s) H+(aq) + H2PO4−(aq)

C. the dissolving of the solid sugar C12H22O11(s) in water

C12H22O11(s) C12H22O11(aq)

H2O(l)

H2O(l)

2H2O(l)


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Equivalents of Electrolytes

An equivalent (Eq) is the amount of an electrolyte

or an ion that provides 1 mole of electrical charge

(+ or −). In solution,

• the charge of the positive ions is always balanced

by the charge of the negative ions.

• the concentrations of electrolytes in intravenous

fluids are expressed in milliequivalents per liter

(mEq/L):

1 Eq = 1000 mEq


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Equivalents, Milliequivalenets

For example, a solution containing

• 25 mEq/L of Na+ and 4 mEq/L of K+ has a total positive charge of 29 mEq/L.

• Cl− as the only anion must have a concentration of 29 mEq/ L.


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Calculating Equivalents

The laboratory tests for a patient indicate a blood calcium level

of 8.8 mEq/L.

A. How many moles of calcium ion are in 0.50 L of blood?

We can then convert equivalents to moles (for Ca2+ there are 2 Eq per mole).

×

×

×


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Calculating Equivalents

The laboratory tests for a patient indicate a blood calcium level of 8.8 mEq/L.

B. If chloride ion is the only other ion present, what is its concentration in mEq/L?

If the concentration of Ca2+ is 8.8 mEq/L, then the concentration of Cl− must be 8.8 mEq/L to balance the charge.

Typical concentrations of electrolytes in blood plasma

• have a charge balance; the total number of positive charges is

equal to the total number of negative charges.

• varies due to the nutritional, electrolyte, and fluid needs of the

patient.

Chemistry Link to Health:Electrolytes in Body Fluids


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Study Check

1. In 1 mole of Fe3+, there are __________.

A. 1 Eq B. 2 Eq C. 3 Eq

2. In 2.5 moles of SO42−, there are __________.

A. 2.5 Eq B. 5.0 Eq C. 1.0 Eq

3. An IV bottle contains NaCl. If the Na+ is 34 mEq/L,

the Cl− is __________.

A. 34 mEq/L B. 0 mEq/L C. 68 mEq/L


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Solution

1. In 1 mole of Fe3+, there are C. 3 Eq

2. In 2.5 moles of SO42−, there are B. 5.0 Eq

2.5 mole SO42− × 2 Eq = 5.0 Eq

1 mole SO42−

3. An IV bottle contains NaCl. If the Na+ is 34 mEq/L,

the Cl− is A. 34 mEq/L


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9.3 Solubility

Gout primarily affects adult men over the age of 40.

Attacks of gout may occur when the concentration of uric acid in blood plasma exceeds its solubility of 7 mg/100 mL of

plasma at 37 °°°°C.

Learning Goal Define solubility; distinguish between an unsaturated and a saturated solution. Identify an ionic compound as soluble or insoluble.


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Solubility

Solubility is

• the maximum amount of solute that dissolves in a specific amount of solvent.

• temperature sensitive for solutes. • expressed as grams of solute in 100 grams

of solvent, usually water.


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Unsaturated Solution

Unsaturated solutions

• contain less than the maximum amount of solute.

• can dissolve more solute.


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Saturated Solution

Saturated solutions

• contain the maximum amount of solute that

can dissolve.

• have undissolved solute at the bottom of the

container.

• contain solute that dissolves as well as solute

that recrystallizes in an equilibrium process.

solute + solvent saturated solution

solute dissolves

solute recrystallizes


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Saturated Solution

More solute can dissolve in an unsaturated solution but not in a saturated solution.


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Study Check

Identify each of the following solutions as saturated or unsaturated.

A. Salt disappears when put in water.

B. Sugar added to a cup of water does not disappear, but sits at the bottom of the cup.


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Solution

Identify each of the following solutions as saturated or unsaturated.

A. Unsaturated: Salt disappears when put in water.

B. Saturated: Sugar added to a cup of water does not disappear, but sits at the bottom of the cup.


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Study Check

At 40 °C, the solubility of KBr is 80 g/100 g of H2O. Identify the following solutions as either saturated or unsaturated. Explain.

A. 60 g KBr added to 100 g of water at 40 °C

B. 200 g KBr added to 200 g of water at 40 °C

C. 25 g KBr added to 50 g of water at 40 °C


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Solution

At 40 °C, the solubility of KBr is 80 g/100 g of H2O. Identify the following solutions as either saturated or unsaturated. Explain.

A. Unsaturated: 60 g KBr/100 g of water at 40 °C is less than the solubility of KBr in water (80 g KBr/100 g water).

B. Saturated: 200 g KBr/200 g of water at 40 °C is greater than the solubility of KBr in water (80 g KBr/100 g water).

C. Unsaturated: 25 g KBr/50 g of water at 40 °C is less than the solubility of KBr in water (80 g KBr/100 g water).


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Effect of Temperature on Solubility

Solubility

• depends on temperature.

• of most solids increases as the temperature increases.

• of gases decreases as the temperature increases. In water, most common solids are more

soluble as the temperature increases.


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Effect of Temperature on Solubility

The solubility of gases decreases as the temperature of the solution increases.


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Study Check

1. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun?

2. Why do fish die in water that is too warm?


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Solution

1. The pressure in a bottle increases as the gas leaves solution when it becomes less soluble at higher temperatures. As pressure increases, the bottle could burst.

2. Because O2 gas is less soluble in warm water, fish cannot obtain the amount of O2

required for their survival.


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Solubility and Pressure

Henry’s law states that

• the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.

• at higher pressures, more gas molecules dissolve in the liquid.

When the pressure of a gas above a solution decreases, the solubility of that gas in the solution also decreases.


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Soluble vs. Insoluble Ionic Compounds

• Only ionic compounds that contain a soluble cation or anion are soluble in water.

• In an insoluble ionic compound, the ionic bonds are too strong for the polar water molecules to break. We can use the solubility rules to predict whether an ionic compound would be expected to dissolve in water.


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Mixing certain aqueous solutions produces insoluble ionic compounds.

Core Chemistry Skill Using Solubility Rules

If an ionic compound contains a combination of a cation and an anion that are not

soluble, that ionic compound is insoluble. For example, combinations of cadmium

and sulfide, iron and sulfide, lead and iodide, and nickel and hydroxide do not

contain any soluble ions. Thus, they form insoluble ionic compounds.


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Sulfates, SO42−, are soluble

unless combined with Ba2+, Pb2+, Ca2+, Sr2+, or Hg2

2+.

Barium sulfate, BaSO4, an insoluble ionic compound, is used to enhance X-rays.

Core Chemistry Skill Using Solubility Rules


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Using Solubility Rules


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Study Check

Predict if the following compounds are soluble or insoluble. Explain why.

A. CdS

B. Na2SO4

C. PbI2

D. Ni(NO3)2


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Solution

Predict if the following compounds are soluble or insoluble.

Explain why.

A. CdS Insoluble; S2− compounds are generally

insoluble.

B. Na2SO4 Soluble; Na+ compounds are always soluble.

C. PbI2 Insoluble; I− is soluble unless combined

with Pb2+.

D. Ni(NO3)2 Soluble; NO3− compounds are always soluble.


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Guide to Writing an Equation for the Formation of a Solid


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Study Check

We can use solubility rules to predict whether a solid, called a

precipitate, forms when two solutions of ionic compounds are mixed.

What precipitate forms when solutions of Pb(NO3)2 and K2SO4

are mixed?


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Solution

We can use solubility rules to predict whether a solid, called

a precipitate, forms when two solutions of ionic compounds are mixed.


are mixed?

STEP 1 Write the ions of the reactants.

Reactants, initial combinations:

Pb2+(aq) NO3−(aq)

K+(aq) SO42−(aq)


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Solution




are mixed?

STEP 2 Write combinations of ions and determine if any are insoluble.

Mixture Product Soluble


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Solution




are mixed?

STEP 3 Write the ionic equation including any solid.

Pb2+(aq) + SO42−(aq) + 2K+(aq) + 2NO3

−(aq) �

PbSO4(s) + 2K+(aq) + 2NO3−(aq)


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Solution




are mixed?

STEP 4 Write the net ionic equation.

Remove the spectator ions.Pb2+(aq) + SO4

2−(aq) + 2K+(aq) + 2NO3−(aq) �

PbSO4(s) + 2K+(aq) + 2NO3−(aq)

Pb2+(aq) + SO42−(aq) � PbSO4(s)


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9.4 Solution Concentrations and Reactions

Suppose we

prepare a solution from 8.00 g of KI (solute) and 42.00 g of water solvent).

Learning Goal Calculate the concentration of a solute in a solution; use concentration units to calculate the amountof solute or solution. Given the volume and concentration of a solution, calculate the amount of another reactant or product in a reaction.


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Solution Concentrations

• The amount of a solute may be expressed in

units of grams, milliliters, or moles.

• The amount of a solution may be expressed in

units of grams, milliliters, or liters.


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Mass Percent (m/m) Concentration

Mass percent (m/m) is

• the concentration by mass of solute in mass of solution.

• the grams of solute in 100 grams of solution.

×

×


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Mass of Solute − Mass Solution

When 42.00 g of water is

added to 8.00 g of KCl, the

mass percent concentration

is 16.0% (m/m).

×

×


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Guide to Calculating Solution Concentration


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Calculating Mass Percent

What is the mass percent of NaOH in a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of H2O?

STEP 1 State the given and needed quantities.

ANALYZE Given Needed

THE 30.0 g NaOH solute mass percent

PROBLEM 30.0 g NaOH + 120.0 g H2O = (m/m)

150.0 g of NaOH solution


THE 30.0 g NaOH solute mass percent

PROBLEM 30.0 g NaOH + 120.0 g H2O = (m/m)

150.0 g of NaOH solution


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Calculating Mass Percent

What is the mass percent of NaOH in a solution prepared

by dissolving 30.0 g of NaOH in 120.0 g of H2O?

STEP 2 Write the concentration expression.

STEP 3 Substitute solute and solution quantities into the expression and calculate.

×

×


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Study Check

A solution is prepared by mixing 15.0 g of Na2CO3

and 235 g of H2O. Calculate the mass percent (m/m)

of the solution.

A. 15.0% (m/m) Na2CO3 solution

B. 6.38% (m/m) Na2CO3 solution

C. 6.00% (m/m) Na2CO3 solution


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Solution

A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g

of H2O. Calculate the mass percent (m/m) of the solution.



ANALYZE Given Need

THE 15.0 g Na2CO3 solute mass percent (m/m)

PROBLEM 15.0 g Na2CO3 + 235 g H2O = 250. g solution

ANALYZE Given Need

THE 15.0 g Na2CO3 solute mass percent (m/m)

PROBLEM 15.0 g Na2CO3 + 235 g H2O = 250. g solution

×


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Solution

A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g

of H2O. Calculate the mass percent (m/m) of the solution.


The answer is C, 6.00% Na2CO3 solution.

×


Karen C. Timberlake


Volume Percent (v/v) Concentration

The volume percent (v/v) is the

• percent volume (mL) of solute (liquid) to volume (mL)

of solution.

• volume of solute (mL) in 100 mL of solution (conversion factor for volume percent)

×


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Mass/Volume Percent

The mass/volume percent (m/v) is the

• percent mass (g) of solute to volume (mL) of solution.

• mass of solute (g) in 100 mL of solution.

(conversion factor for mass/volume percent)

×

Core Chemistry Skill Calculating Concentration


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Study Check

Write two conversion factors for each solution.

A. 8.50% (m/m) NaOH

B. 5.75% (v/v) ethanol

C. 4.8% (m/v) HCl


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Solution

A. 8.50% (m/m) NaOH

B. 5.75% (v/v) ethanol

C. 4.8% (m/v) HCl


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Molarity

Molarity (moles of solute/liter of solution) is defined as the moles of solute per volume (L) of solution.

A 1.0 M solution of NaCl is defined as


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Molarity Calculations

What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH?


1 mole NaOH = 40.00 g NaOH

ANALYZE Given Need

THE 6.00 g NaOH solute molarity (mole/L)

PROBLEM 0.500 L of NaOH solution

ANALYZE Given Need

THE 6.00 g NaOH solute molarity (mole/L)

PROBLEM 0.500 L of NaOH solution

×


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Molarity Calculations

What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH?




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Study Check

What is the molarity of 0.225 L of a KNO3

solution containing 34.8 g of KNO3?

A. 0.344 M

B. 1.53 M

C. 15.5 M


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Solution

What is the molarity of 0.225 L of a KNO3 solution containing 34.8 g of KNO3?


1 mole of KNO3 = 101.11 g KNO3

ANALYZE Given Need

THE 34.8 g KNO3 solute molarity (mole/L)

PROBLEM 0.225 L of KNO3 solution

ANALYZE Given Need

THE 34.8 g KNO3 solute molarity (mole/L)

PROBLEM 0.225 L of KNO3 solution

×


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Solution

What is the molarity of 0.225 L of a KNO3 solution

containing 34.8 g of KNO3?



The answer is B, 1.53 M KNO3.


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Molarity as a Conversion Factor

The units of molarity are used as conversion factors in

calculations with solutions.

Molarity Equality

3.5 M HCl 1 L solution = 3.5 moles of HCl

Written as Conversion Factors


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Conversion Factors, Concentrations

Core Chemistry Skill Using Concentration as a Conversion Factor


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Guide to Using Concentration to Calculate Mass or Volume


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Study Check

How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution?

A. 10.5 g of NaOH

B. 75.0 g of NaOH

C.536 g of NaOH


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Solution

How many grams of NaOH are needed to prepare 75.0 g of

14.0% (m/m) NaOH solution?


STEP 2 Write a plan to calculate mass or volume.grams of grams of solution solute

ANALYZE Given Need

THE 75.0 grams of 14.0% grams NaOH

PROBLEM (m/m) NaOH solution

ANALYZE Given Need

THE 75.0 grams of 14.0% grams NaOH

PROBLEM (m/m) NaOH solution

Percent Percent

(m/m)


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Solution

How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution?

STEP 3 Write equalities and conversion factors.

14.0 g NaOH = 100 g of solution

STEP 4 Set up the problem to calculate mass or volume.

The answer is A, 10.5 g NaOH.

×


Karen C. Timberlake


Study Check

How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution?

A. 20.0 g of AlCl3

B. 16.7 g of AlCl3

C. 2.50 g of AlCl3


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Solution

How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution?


STEP 2 Write a plan to calculate mass or volume.liters of grams of solution solute

ANALYZE Given Need

THE 125 mL (0.125 L) solution grams AlCl3

PROBLEM 0.150 M solution

ANALYZE Given Need

THE 125 mL (0.125 L) solution grams AlCl3

PROBLEM 0.150 M solution

Molarity


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Solution

How many grams of AlCl3 are needed to prepare 125 mL of a

0.150 M solution?

STEP 3 Write equalities and conversion factors.

0.150 moles AlCl3 = 1 L of solution

1 mole AlCl3 = 133.33 g AlCl3


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Solution

How many grams of AlCl3 are needed to prepare 125 mL of a

0.150 M solution?

STEP 4 Set up the problem to calculate mass or volume.

Answer is C, 2.50 g AlCl3.

××


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Chemical Reactions in Solution

When chemical reactions involve aqueous solutions, we use

• the balanced chemical equation, • the molarity, and• the volumeto determine the moles or grams of the reactants or products.

Core Chemistry Skill Calculating the Quantity of a Reactant or Product for a Chemical Reaction in Solution


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Guide to Calculations Involving Solutions in Chemical Reactions


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Zinc reacts with HCl to produce hydrogen gas, H2 ,

and ZnCl2.

Zn(s) + 2HCl(aq)� H2(g) + ZnCl2(aq)

How many liters of a 1.50 M HCl solution completely

react with 5.32 g of zinc?

Study Check


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Solution

Zinc reacts with HCl to produce hydrogen gas, H2 , and ZnCl2.


How many liters of a 1.50 M HCl solution completely react with 5.32 g of zinc?


ANALYZE Given Need

THE 5.32 g Zn

PROBLEM 1.50 M HCl solution liters of HCl

Equation solution


ANALYZE Given Need

THE 5.32 g Zn

PROBLEM 1.50 M HCl solution liters of HCl

Equation solution



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Solution

Zinc reacts with HCl to produce hydrogen gas, H2 , and ZnCl2.



STEP 2 Write a plan to calculate the needed quantity.

moles moles volumeof of ofzinc HCl HCl

Mole–mole

factorMolarity


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Solution

Zinc reacts with HCl to produce hydrogen gas, H2, and ZnCl2.



STEP 3 Write equalities and conversion factors including mole–mole and concentration factors.


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Solution

Zinc reacts with HCl to produce hydrogen gas, H2, and ZnCl2.



STEP 4 Set up the problem to calculate needed quantity.

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Concept Map for Chemical Equations


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9.5 Dilution of Solutions

In a dilution, a solvent, usually water, is added to a solution,

which increases its volume and decreases the concentration of the solution. Making orange juice from concentrate is an example of a dilution.

Learning Goal Describe the dilution of a solution; calculate the

unknown concentration or volume when a solution is diluted.


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Dilution

In a dilution,

• water is added.

• the volume of the solution increases.

• concentration decreases.

• the mass of solute in the solution remains the same.


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Solute Concentrations

In the initial and diluted solution,

• the moles of solute are the same.

• the concentrations and volumes are related by the following equation:

C1V1 = C2V2

initial diluted

This equality is written in terms of the concentration, C, and the volume, V. The concentration, C, may be percent concentration or molarity.


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Dilution of a Solution

When water is added to a concentrated solution, there is no

change in the number of particles. The solute particles spread out as the volume of the solution increases.


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Guide to Calculating Dilution Quantities


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Dilution: Molarity

What is the final concentration when 0.50 L of 6.0 M HCl

solution is diluted to a final volume of 1.0 L?

STEP 1 Prepare a table of the concentrations and volumes of the solutions.

ANALYZE Given Need Know Predict

THE C1 = 6.0 M HCl C2 = ? M HCl C decreases

PROBLEM V1 = 0.50 L V increases

V2 = 1.0 L


THE C1 = 6.0 M HCl C2 = ? M HCl C decreases

PROBLEM V1 = 0.50 L V increases

V2 = 1.0 L


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Dilution: Molarity

What is the final concentration when 0.50 L of 6.0 M HCl

solution is diluted to a final volume of 1.0 L?

STEP 2 Rearrange the dilution expression to solve for the unknown quantity.

STEP 3 Substitute the known quantities into the dilution expression and calculate.

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Study Check

What volume of a 2.00% (m/v) HCl solution can be

prepared by diluting 25.0 mL of 14.0% (m/v) HCl

solution?


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Solution

What volume of a 2.00% (m/v) HCl solution can be prepared

by diluting 25.0 mL of 14.0% (m/v) HCl solution?



THE C1 = 14.0% (m/v) V2 = ? M HCl C decreases

PROBLEM C2 = 2.00% (m/v)

V1 = 25.0 mL V increases


THE C1 = 14.0% (m/v) V2 = ? M HCl C decreases

PROBLEM C2 = 2.00% (m/v)



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Solution

What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?



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Study Check

What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOHto 60.0 mL?


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Solution

What is the percent (% m/v) of a solution prepared by diluting

10.0 mL of 9.00% NaOH to 60.0 mL?



THE C1 = 9.00% (m/v) C2 = ? M NaOH C decreases

PROBLEM V1 = 10.0 mL



THE C1 = 9.00% (m/v) C2 = ? M NaOH C decreases

PROBLEM V1 = 10.0 mL



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Solution

What is the percent (% m/v) of a solution prepared by diluting

10.0 mL of 9.00% NaOH to 60.0 mL?



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Study Check

What is the molarity (M) of a solution prepared by

diluting 0.180 L of 0.600 M HNO3 to 0.540 L?


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Solution

What is the molarity (M) of a solution prepared by diluting

0.180 L of 0.600 M HNO3 to 0.540 L?



THE M1 = 0.600 M M2 = ? M HNO3 M decreases

PROBLEM V1 = 0.180 L

V2 = 0.540 L V increases


THE M1 = 0.600 M M2 = ? M HNO3 M decreases

PROBLEM V1 = 0.180 L

V2 = 0.540 L V increases


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Solution

What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?



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9.6 Properties of Solutions

Ethylene glycol is

added to a radiator

to form an aqueous

solution that has a

lower freezing point

than water.

Learning Goal Identify a mixture as a solution, a colloid, or a suspension. Describe how the number of particles in a solution affects the freezing point, the boiling point, and the osmotic pressure of a solution.


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Solutions or Colloids

Solutions

• are transparent.

• do not separate.

• contain small particles, ions, or molecules that cannot be filtered and pass through semipermeable membranes.

Colloids

• have medium-sized particles.

• cannot be filtered.

• can be separated by semipermeable membranes.


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Examples of Colloids


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Suspensions

Suspensions

• are heterogeneous, nonuniform mixtures.

• have very large particles that settle out of solution.

• can be filtered.

• must be stirred to stay suspended.

Examples include blood platelets, muddy water, and

calamine lotion.


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Solutions, Colloids, and Suspensions


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Solutions, Colloids, and Suspensions

Properties of different

types of mixtures: (a) suspensions settle out; (b) suspensions are separated by a filter; (c) solution particles go through a semipermeablemembrane, but colloids and suspensions do not.


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Study Check

A mixture that has solute particles that do not settle

out but are too large to pass through a

semipermeable membrane is called a _______.

A. solution

B. colloid

C. suspension


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Solution

A mixture that has solute particles that do not settle

out but are too large to pass through a semipermeable

membrane is called a ______.

B. colloid

When a solute is added to water, the physical properties

change and the

• vapor pressure above the solution decreases.

• boiling point of the solution increases.

• freezing point of the solution decreases.

These types of properties are called colligative properties; they depend only on the concentration of solute particles in the solution.

Freezing Point Lowering, Boiling Point Elevation


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Vapor Pressure Lowering

Increasing the concentration of solute particles in the solution

• decreases the number of solvent particles at the surface of the solution.

• prevents some of the solvent particles from leaving the solution.


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Boiling Point Elevation

Increasing the concentration of nonvolatile solute

particles in the solution raises the boiling point of

the solution, and

• more solute particles in the solution lowers the

vapor pressure.

• the solution boils at a higher temperature than

the normal boiling point.


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Freezing Point Lowering

Adding salt to an icy road when temperatures drop below freezing

Adding ethylene glycol to water increases the number of hydrogen bonds that form in the solution, lowering the freezing point and raising the boiling point of the solution.

• allows the particles of salt to mix with the water.

• lowers the freezing point of the ice.


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Particles in Solution

A solute that is a nonelectrolyte dissolves as molecules,

whereas a solute that is a strong electrolyte dissolves entirely as ions.

• The solute in antifreeze, which is ethylene glycol, C2H6O2, is a nonelectrolyte and dissolves as molecules in water.

1 mole C2H6O2(l) = 1 mole C2H6O2(aq)

• The solute NaCl, a strong electrolyte, dissolves as ions in water.

NaCl(s) � Na+(aq) + Cl− (aq)

1 mole NaCl(s) = 1 mole Na+(aq) + 1 mole Cl− (aq)

1 mole NaCl(s) = 2 moles of particles in solution


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Particles in Solution

The effect of solute particles on boiling points and freezing points is summarized below:

Guide to Calculating Boiling Point Elevation, Freezing Point Lowering

Core Chemistry Skill Calculating the Boiling Point/Freezing Point

of a Solution


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Calculating Freezing Point Lowering

In the northeastern United States during freezing

temperatures, CaCl2 is spread on icy highways to melt the ice. Calculate the freezing point of a solution containing 0.50 mole of CaCl2 in 1 kg of water.



THE 0.50 mole of CaCl2 freezing point of solution

PROBLEM 1 kg of water


THE 0.50 mole of CaCl2 freezing point of solution

PROBLEM 1 kg of water


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STEP 2 Determine the number of moles of solute particles.

CaCl2(s) � Ca2+(aq) + 2Cl−(aq)

1 mole of CaCl2 = 3 moles of solute particles


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STEP 3 Determine the temperature change using the moles of solute particles and the degrees Celsius change per mole of particles.

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Osmosis

In osmosis,

• water (solvent) flows from a lower to a higher solute concentration.

• the level of the solution with the higher solute concentration rises.

• the concentrations of the two solutions become equal with time.

Water flows into the solution with a higher solute concentration until the flow of water becomes equal in both directions.


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Osmotic Pressure

Osmotic pressure is

• equal to the pressure that would prevent the flow of additional water into the more concentrated solution.

• greater as the number of dissolved particles in the solution increases.


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Reverse Osmosis

In a process called reverse osmosis,

• a pressure greater than the osmotic pressure is applied to a solution, forcing it through a purification membrane.

• the flow of water is reversed because water flows from an area of lower to higher water concentration, leaving behind the molecules and ions in solution.

Reverse osmosis, used in desalination plants to obtain pure water from sea (salt) water, requires a large amount of energy.


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Isotonic Solutions

Because cell membranes in biological systems are

semipermeable, osmosis is an ongoing process.

The solutes in body solutions such as blood, tissue fluids, lymph, and plasma all exert osmotic pressure. Most IV solutions used in hospitals are isotonic solutions, which

• exert the same osmotic pressure as body fluids such as red blood cells (RBCs).

• include a 5.0% (m/v) glucose or 0.90% (m/v) NaClisotonic solution.


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Hypotonic and Hypertonic Solutions

(a) In an isotonic solution, a red blood cell retains its normal volume. (b) Hemolysis: In a

hypotonic solution, water flows into a red blood cell, causing it to swell and burst. (c)

Crenation: In a hypertonic solution, water leaves the red blood cell, causing it to shrink.


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Hypotonic Solution

A hypotonic solution

• has a lower solute concentration than red blood cells.

• means water flows into cells by osmosis.

The increase in fluid causes the cells to swell and burst, a condition called hemolysis.

(b) Hemolysis: In a hypotonic

solution, water flows into a red blood cell, causing it to swell and burst.


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Hypertonic Solution

A hypertonic solution

• has a higher solute concentration than RBCs.

• Involves water going out of the cells by osmosis.

• causes crenation: RBCs shrink in size.

(c) Crenation: In a hypertonic

solution, water leaves the red blood cell, causing it to shrink.


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Dialysis

In dialysis,

• solvent and small solute particles pass through an artificial

membrane.

• large particles are retained inside.

• waste particles such as urea from blood are removed using

hemodialysis (artificial kidney).

During dialysis, waste products and excess water are removed from the blood.


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Study Check

Indicate if each of the following solutions is

A. isotonic. B. hypotonic. C. hypertonic.

1. ____ 2% NaCl solution

2. ____ 1% glucose solution

3. ____ 0.5% NaCl solution



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Solution

Indicate if each of the following solutions is

A. isotonic. B. hypotonic. C. hypertonic.

1. _C__ 2% NaCl solution

2. _B__ 1% glucose solution

3. _B__ 0.5% NaCl solution

4. _A__ 5% glucose solution


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Study Check

When placed in each of the following, indicate if a red

blood cell will

A. not change. B. undergo hemolysis. C. undergo crenation.



3. ____ 0.5% NaCl solution

4. ____ 2% NaCl solution


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Solution

When placed in each of the following, indicate if a red

blood cell will

A. not change. B. undergo hemolysis. C. undergo crenation.

1. _A__ 5% glucose solution

2. _B__ 1% glucose solution

3. _B__ 0.5% NaCl solution

4. _C__ 2% NaCl solution


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Concept Map Solutions

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